Question

【高等数学数列求和问题求解】解答过程写纸上，必给好评，谢谢！

Answer 1

解：
(1)
(a3+a8)-(a2+a7)=2d=-29-(-23)=-6
d=-3
a2+a7=-23
2a1+7d=-23
a1=(-23-7d)/2=[-23-7×(-3)]/2=-1
an=a1+(n-1)d=-1+(-3)(n-1)=-3n+2
数列{an}的通项公式为an=-3n+2
(2)
an+bn=1·qⁿ⁻¹=qⁿ⁻¹
bn=qⁿ⁻¹-an=qⁿ⁻¹-(-3n+2)=qⁿ⁻¹+3n-2
q=1时，bn=1+3n-2=3n-1
Sn=b1+b2+...+bn
=3(1+2+...+n)-n
=3n(n+1)/2 -n
=n(3n+1)/2
q≠0且q≠1时，
Sn=1·(qⁿ-1)/(q-1) +3(1+2+...+n) -2n
=(qⁿ-1)/(q-1) +3n(n+1)/2 -2n
=(qⁿ-1)/(q-1) +n(3n-1)/2

Answer 2

a2+a7 = 2a1+7d = -23
a3+a8 = 2a1+9d = -29
联立解得 d = -3， a1 = -1
则 an = -1-3(n-1) = 2-3n
{an} 前 n 项和 是 Pn = (1/2)n(1-3n)
a1+b1 = 2-3+b1 = 1, b1 = 2
a2+b2 = 2-6+2q = q, q = 4
{an+bn} 前 n 项和 是 Qn = q^(n-1) = 4^(n-1)
则 {bn} 前 n 项和 是 Sn= Qn - Pn = 4^(n-1) + (1/2)n(3n-1)